If the activity of ${}^{108}$$Ag$ is $3$ micro curie, the number of atoms present in it are ( $\lambda =$ 0.005 sec${}^{-1}$)

$\lambda =0.005se{c}^{-1}$
$1$ Micro curie $=3.7\times 10^{4}dps$
Assuming all $Ag$ atoms to be radioactive, and each showing
Let no of atoms present $=1dpsno$
Given:
$(\frac{dN}{dt})activity=\lambda N_0\Rightarrow 3.7\times 3\times 10^{4}dps=0.005\times N_0$
                                        $\Rightarrow N_0=2.2\times 10^{7}atoms$